Mathematical Induction...October 1, 2010 Brian Forrest Mathematical Induction Principle of...

Mathematical Induction

Brian Forrest

October 1, 2010

Brian Forrest Mathematical Induction

Principle of Mathematical Induction

Axiom: (Principle of Mathematical Induction)

Suppose that S ! N is such that

! I1) 1 " S

! I2) if n " S , then n + 1 " S .

then S = N.

Axiom: (Principle of Mathematical Induction)Suppose that S ! N is such that

! I1) 1 " S

! I2) if n " S , then n + 1 " S .

then S = N.

! I1) 1 " S

! I2) if n " S ,

then n + 1 " S .

then S = N.

! I1) 1 " S

! I2) if n " S , then n + 1 " S .

then S = N.

! I1) 1 " S

! I2) if n " S , then n + 1 " S .

then S = N.

Question: Why is the Principle of Mathematical Inductionreasonable?

Test: Suppose S satisfies I1) and I2). Show that 7 " S .

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

Suppose S satisfies I1) and I2). Show that 7 " S .

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

Test: Suppose S satisfies I1) and I2).

Show that 7 " S .

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

! 1 " S I1)

! # 1 + 1 = 2 " S I2)

! # 2 + 1 = 3 " S I2)

! # 3 + 1 = 4 " S I2)

! # 4 + 1 = 5 " S I2)

! # 5 + 1 = 6 " S I2)

! # 6 + 1 = 7 " S I2)

Question:

How is the Principle of Mathematical Induction useful?

• Leads to Proof by Induction or Induction for short.

Question: How is the Principle of Mathematical Induction useful?

• Leads to Proof by Induction

or Induction for short.

Proofs by Induction

Question:

How does a proof by induction work?

1) The setup:

• Assign to each natural number n " N the statement P(n) thatwe want to prove.

Example: P(n) is the statement that:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

• Let S be the set of all natural numbers n such that thestatement P(n) is true.

2) The Goal: Show that S = N.

Proofs by Induction

Question: How does a proof by induction work?

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

• Assign to each natural number n " N

the statement P(n) thatwe want to prove.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

Example:

P(n) is the statement that:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

2) The Goal:

Show that S = N.

Proofs by Induction

1) The setup:

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Proofs by Induction

3) The Method:

• Step 0)

Clearly state what P(n) represents.

• Step 1) Initial Step:

Show that P(1) is true.

•Step 2) Inductive Step:

Assume that P(k) holds true for some k (this is called theinduction hypothesis) and then use this assumption to show thatP(k + 1) also holds.

Proofs by Induction

3) The Method:

• Step 0)

Proofs by Induction

3) The Method:

• Step 0)

Proofs by Induction

3) The Method:

• Step 0)

Proofs by Induction

3) The Method:

• Step 0)

Proofs by Induction

3) The Method:

• Step 0)

Proofs by Induction

3) The Method:

• Step 0)

Assume that P(k) holds true for some k

(this is called theinduction hypothesis) and then use this assumption to show thatP(k + 1) also holds.

Proofs by Induction

3) The Method:

• Step 0)

Assume that P(k) holds true for some k (this is called theinduction hypothesis)

and then use this assumption to show thatP(k + 1) also holds.

Proofs by Induction

3) The Method:

• Step 0)

Example

Example:

Prove that

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Step 0): Let P(n) be the statement that

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Example

Example: Prove that

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Example

Example: Prove that

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

Initial Step

Step 1): Show that P(1) holds.

=1(1 + 1)

This establishes P(1).

Initial Step

=1(1 + 1)

Initial Step

=1(1 + 1)

Initial Step

=1(1 + 1)

Initial Step

=1(1 + 1)

Inductive Step

Step 2):

Assume that P(k) holds. Consider

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

This shows that P(k + 1) also holds.

Inductive Step

Step 2): Assume that P(k) holds.

Consider

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

Step 2): Assume that P(k) holds. Consider

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Inductive Step

% + (k + 1)

=k(k + 1)

2+ (k + 1) (by P(k))

'(k + 1)

=(k + 2)(k + 1)

=(k + 1)((k + 1) + 1)

Conclusion

We have shown that:

! P(1) holds.

! If P(k) holds, so does P(k + 1).

We conclude by induction that

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Conclusion

We have shown that:

! P(1) holds.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Conclusion

We have shown that:

! P(1) holds.

! If P(k) holds,

so does P(k + 1).

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Conclusion

We have shown that:

! P(1) holds.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Conclusion

We have shown that:

! P(1) holds.

j = 1 + 2 + 3 + · · · + n =n(n + 1)

for all n " N.

Shaggy Dogs

Important Note:

Care must be taken with induction proofs!

Problem: Prove that all dogs are shaggy!

Shaggy Dogs

Important Note: Care must be taken with induction proofs!

Shaggy Dogs

Problem:

Prove that all dogs are shaggy!

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof:

Assume that P(k)=everygroup of k dogs is shaggy.Consider a group of k + 1 dogs.Remove one dog from the group.The new group contains k dogsso all are shaggy.

Next return the original dog to

the group and remove one shaggy

dog.Again we have a group of k dogs, so all of these dogs are shaggy. .Therefore, the original dog is shaggy.Hence the group of k + 1 dogs are all shaggy. By induction we haveproved that all dogs are shaggy!!!

Question: What’s wrong with this proof?

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof: Assume that P(k)=everygroup of k dogs is shaggy.

Consider a group of k + 1 dogs.Remove one dog from the group.The new group contains k dogsso all are shaggy.

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof: Assume that P(k)=everygroup of k dogs is shaggy.Consider a group of k + 1 dogs.

Remove one dog from the group.The new group contains k dogsso all are shaggy.

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof: Assume that P(k)=everygroup of k dogs is shaggy.Consider a group of k + 1 dogs.Remove one dog from the group.

The new group contains k dogsso all are shaggy.

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof: Assume that P(k)=everygroup of k dogs is shaggy.Consider a group of k + 1 dogs.Remove one dog from the group.The new group contains k dogs

so all are shaggy.

Shaggy Dogs

=unknown =shaggy

=not shaggy

Proof: Assume that P(k)=everygroup of k dogs is shaggy.Consider a group of k + 1 dogs.Remove one dog from the group.The new group contains k dogsso all are shaggy.

Shaggy Dogs

=unknown =shaggy

=not shaggy

the group

and remove one shaggy

Shaggy Dogs

=unknown =shaggy

=not shaggy

Again we have a group of k dogs, so all of these dogs are shaggy. .Therefore, the original dog is shaggy.Hence the group of k + 1 dogs are all shaggy. By induction we haveproved that all dogs are shaggy!!!

Shaggy Dogs

=unknown =shaggy

=not shaggy

dog.Again we have a group of k dogs,

so all of these dogs are shaggy. .Therefore, the original dog is shaggy.Hence the group of k + 1 dogs are all shaggy. By induction we haveproved that all dogs are shaggy!!!

Shaggy Dogs

=unknown =shaggy

=not shaggy

dog.Again we have a group of k dogs, so all of these dogs are shaggy.

.Therefore, the original dog is shaggy.Hence the group of k + 1 dogs are all shaggy. By induction we haveproved that all dogs are shaggy!!!

Shaggy Dogs

=unknown =shaggy

=not shaggy

dog.Again we have a group of k dogs, so all of these dogs are shaggy. .Therefore, the original dog is shaggy.

Hence the group of k + 1 dogs are all shaggy. By induction we haveproved that all dogs are shaggy!!!

Shaggy Dogs

=unknown =shaggy

=not shaggy

dog.Again we have a group of k dogs, so all of these dogs are shaggy. .Therefore, the original dog is shaggy.Hence the group of k + 1 dogs are all shaggy.

By induction we haveproved that all dogs are shaggy!!!

Shaggy Dogs

=unknown =shaggy

=not shaggy

Shaggy Dogs

=unknown =shaggy

=not shaggy

Question:

What’s wrong with this proof?

Shaggy Dogs

=unknown =shaggy

=not shaggy

Tower of Hanoi

Problem: (Tower of Hanoi)

You are given three pegs. On one of thepegs is a tower made up of n rings placed on top of one another so thatas you move down the tower each successive ring has a larger diameterthan the previous ring. The object of this puzzle is to reconstruct thetower on one of the other pegs by moving one ring at a time, from onepeg to another, in such a manner that you never have a ring aboveany smaller ring on any of the three pegs.

Task: Prove that for any n " N, if you begin with n rings, then thepuzzle can be completed in 2n $ 1 moves.

Moreover, prove that for each n this is the minimum number of movesnecessary to complete the task.

Question: What is P(n)?

Tower of Hanoi

Problem: (Tower of Hanoi) You are given three pegs. On one of thepegs is a tower made up of n rings

placed on top of one another so thatas you move down the tower each successive ring has a larger diameterthan the previous ring. The object of this puzzle is to reconstruct thetower on one of the other pegs by moving one ring at a time, from onepeg to another, in such a manner that you never have a ring aboveany smaller ring on any of the three pegs.

Tower of Hanoi

Problem: (Tower of Hanoi) You are given three pegs. On one of thepegs is a tower made up of n rings placed on top of one another so thatas you move down the tower each successive ring has a larger diameterthan the previous ring.

The object of this puzzle is to reconstruct thetower on one of the other pegs by moving one ring at a time, from onepeg to another, in such a manner that you never have a ring aboveany smaller ring on any of the three pegs.

Tower of Hanoi

Problem: (Tower of Hanoi) You are given three pegs. On one of thepegs is a tower made up of n rings placed on top of one another so thatas you move down the tower each successive ring has a larger diameterthan the previous ring. The object of this puzzle is to reconstruct thetower on one of the other pegs by moving one ring at a time, from onepeg to another, in such a manner that you never have a ring aboveany smaller ring on any of the three pegs.

Tower of Hanoi

Prove that for any n " N, if you begin with n rings, then thepuzzle can be completed in 2n $ 1 moves.

Tower of Hanoi

Question:

What is P(n)?

Tower of Hanoi

Introduction to Sequences:

Introduction to Sequences

Brian Forrest

October 1, 2010

Brian Forrest Introduction to Sequences

What is a sequence?

Informally, a sequence is an ordered list of real numbers.

For example a phone number 519-555-1234 is a finite sequence of1-digit numbers

5! 1! 9! 5! 5! 5! 1! 2! 3! 4.

Note: Order is important.

For us sequences will be infinite ordered lists of real numbers ofthe form {a1, a2, a3, . . . , an, . . .}.

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

1! 9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9!

5! 5! 5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5!

5! 5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5!

5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5! 5!

1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5! 5! 1!

2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2!

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3!

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3! 4

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

5! 1! 9! 5! 5! 5! 1! 2! 3! 4.

What is a sequence?

More formally:

Definition: A sequence is a function f : N" R.

For n # N, we call f (n) the nth term of the sequence and n iscalled the index.

Notation: We often write an, or bn, or xn instead of f (n) todenote the nth term.

What is a sequence?

More formally:

What is a sequence?

More formally:

What is a sequence?

More formally:

Notation:

We often write an, or bn, or xn instead of f (n) todenote the nth term.

What is a sequence?

More formally:

Notation: We often write an,

or bn, or xn instead of f (n) todenote the nth term.

What is a sequence?

More formally:

Notation: We often write an, or bn,

or xn instead of f (n) todenote the nth term.

What is a sequence?

More formally:

Notation: We often write an, or bn, or xn

instead of f (n) todenote the nth term.

What is a sequence?

More formally:

Ways to specify a sequence

Ways to specify a sequence:

1) Give the explicit function:

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

! f (n) = 1n

! an = 1n

! {1, 12 , 1

3 , . . . , 1n , . . .}

! { 1n}

2) Recursion:

Future terms in a sequence are defined from existingterms.

Example: Let a1 = 1 and an+1 = 11+an

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

Exercise: Can you find an explicit formula for an?

2) Recursion: Future terms in a sequence are defined from existingterms.

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

Exercise:

Can you find an explicit formula for an?

What is a5?

! a1 = 1

! a2 = 11+a1

= 11+1 = 1

! a3 = 11+a2

= 11+1

! a4 = 11+a3

= 11+2

! a5 = 11+a4

= 11+3

Subsequences: New sequences from old

Definition: Given a sequence {an} and a sequence {nk} # N suchthat

n1 < n2 < n2 . . . < nk . . .

definebk = ank .

The sequence {bk} = {an1 , an2 , an3 , . . . , an1 , . . .} is called asubsequence of {an} .

Example: Let an = 1n and nk = 2k. Then

bk = ank = a2k =1

n1 < n2 < n2 . . . < nk . . .

definebk = ank .

bk = ank = a2k =1

n1 < n2 < n2 . . . < nk . . .

definebk = ank .

bk = ank = a2k =1

n1 < n2 < n2 . . . < nk . . .

definebk = ank .

Example: Let an = 1n and nk = 2k.

bk = ank = a2k =1

n1 < n2 < n2 . . . < nk . . .

definebk = ank .

bk = ank = a2k =1

={a ,a ,a ,a ,...,a ,...}1

2 3 4 2k

={a ,a ,a ,a ,...,a ,...}1

2 3 4 2k

={a ,a ,a ,a ,...,a ,...}1

2 3 4 2k

a ,a ,

b ,b ,1

={a ,a ,a ,a ,...,a ,...}1

2 3 4 2k

2ka ,a ,

b ,b ,1

...,b ,

...,a ,

={a ,a ,a ,a ,...,a ,...}1

2 3 4 2k

2ka ,a ,

b ,b ,1

...,b ,

...,a ,

k ...}

Example: Given {an} and j # N $ {0},

bk = ak+j = {a1+j , a2+j , a3+j , . . .}.

Note: Such a subsequence is called a tail of {an}.

Example: Given {an} and j # N $ {0}, let

bk = ak+j = {a1+j , a2+j , a3+j , . . .}.

Example: Given {an} and j # N $ {0}, let

bk = ak+j = {a1+j , a2+j , a3+j , . . .}.

Graphical Representation of a Sequence

10 20 30 40

1) Standard 2-dimensional graph.

Example: { 1n}.

10 20 30 40

f(n)=1/n

1) Standard 2-dimensional graph.

Example: { 1n}.

0 0.2 0.4 0.6 0.8 1

2) 1-dimensional plot.

Example: { 1n}.

0 0.2 0.4 0.6 0.8 1

2) 1-dimensional plot.

Example: { 1n}.

2 3 4 2k-1 2k

Example: an = (!1)n+1

% {1, ! 1, 1, ! 1, . . . 1, ! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1,

! 1, 1, ! 1, . . . 1, ! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1,

1, ! 1, . . . 1, ! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1, 1,

! 1, . . . 1, ! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1, 1, ! 1,

. . . 1, ! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1, 1, ! 1, . . . 1,

! 1, . . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1, 1, ! 1, . . . 1, ! 1,

. . .}

2 3 4 2k-1 2k

Example: an = (!1)n+1 % {1, ! 1, 1, ! 1, . . . 1, ! 1, . . .}

Least Upper Bound Property:

Least Upper Bound Property

Brian Forrest

September 5, 2010

Brian Forrest Least Upper Bound Property

Bounded Sets

Definition:

Let S ! R1. ! is an upper bound for S if x " ! for all x # S .

2. " is an lower bound for S if " " x for all x # S .

Definition::Let S ! R1. S is bounded above if S has an upper bound !.

2. S is bounded below if S has an lower bound ".

3. S is bounded if S is bound above and bounded below.

Bounded Sets

Definition: Let S ! R

1. ! is an upper bound for S if x " ! for all x # S .

Bounded Sets

Definition: Let S ! R1. ! is an upper bound for S

if x " ! for all x # S .

Bounded Sets

Definition: Let S ! R1. ! is an upper bound for S if x " ! for all x # S .

Bounded Sets

2. " is an lower bound for S

if " " x for all x # S .

Bounded Sets

Definition::

Let S ! R1. S is bounded above if S has an upper bound !.

Bounded Sets

Definition::Let S ! R

1. S is bounded above if S has an upper bound !.

Bounded Sets

Definition::Let S ! R1. S is bounded above

if S has an upper bound !.

Bounded Sets

2. S is bounded below

if S has an lower bound ".

Bounded Sets

3. S is bounded

if S is bound above and bounded below.

Bounded Sets

3. S is bounded if S is bound above

and bounded below.

Bounded Sets

Is N bounded?

Problem:

Is N bounded?

N is bounded below by 1.

Problem: Is N bounded above?

Is N bounded?

Problem: Is N bounded?

Is N bounded?

Problem:

Is N bounded above?

Is N bounded?

Argument 1:

N is bounded above because it has no largestelement!

False! [0, 1) is bounded above, but it has no largest element.

Is N bounded?

Argument 1: N is bounded above because it has no largestelement!

Is N bounded?

False!

[0, 1) is bounded above, but it has no largest element.

Is N bounded?

False! [0, 1) is bounded above,

but it has no largest element.

Is N bounded?

Argument 2:

Assume ! # R is an upper bound for N. Write

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N! m + 1 > !.

This shows that no such ! could be an upper bound.

Is N bounded?

Argument 2: Assume ! # R is an upper bound for N.

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N! m + 1 > !.

Is N bounded?

Argument 2: Assume ! # R is an upper bound for N. Write

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N! m + 1 > !.

Is N bounded?

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N

! m + 1 > !.

Is N bounded?

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N! m + 1 > !.

Is N bounded?

! = m.a1a2a3 · · ·

where m # N.

! m + 1 # N! m + 1 > !.

Is N bounded?

Problem:

How do we know that every real number has a decimalexpansion?

Fact: The existence of a decimal expansion is essentiallyequivalent to N being bounded.

Is N bounded?

Problem: How do we know that every real number has a decimalexpansion?

Is N bounded?

Problem: How do we know that every real number has a decimalexpansion?

[0, 1)

Let S = [0, 1).

! S is bounded above by 2.

! S is also bounded above by 4.

! S has infinitely many upper bounds.

! 1 is a special upper bound for S .

! 1 is the smallest or least upper bound for S .

[0, 1)

Let S = [0, 1).

[0, 1)

0 1 2 4

Let S = [0, 1).

[0, 1)

0 1 2 4

Let S = [0, 1).

[0, 1)

0 1 2 4

Let S = [0, 1).

[0, 1)

0 1 2 4

Let S = [0, 1).

[0, 1)

Let S = [0, 1).

! S is bounded below by $1.

! S is also bounded below by $2.

! S has infinitely many lower bounds.

! 0 is a special lower bound for S .

! 0 is the largest or greatest lower bound for S .

[0, 1)

Let S = [0, 1).

[0, 1)

0 1-1-2

Let S = [0, 1).

[0, 1)

0 1-1-2

Let S = [0, 1).

[0, 1)

0 1-1-2

Let S = [0, 1).

[0, 1)

0 1-1-2

Let S = [0, 1).

Least Upper Bound

Definition:

We say that ! # R is the least upper bound for a setS ! R if

1. ! is an upper bound for S , and

2. if # is an upper bound for S , then ! " # .

If a set S has a least upper bound , then we denote it by lub(S).

The least upper bound of S is often called the supremum of S ,denoted by sup(S).

Least Upper Bound

Definition: We say that ! # R is the least upper bound for a setS ! R if

Least Upper Bound

2. if # is an upper bound for S ,

then ! " # .

Least Upper Bound

If a set S has a least upper bound

, then we denote it by lub(S).

Least Upper Bound

The least upper bound of S is often called the supremum of S ,

denoted by sup(S).

Least Upper Bound

Greatest Lower Bound

Definition:

We say that " # R is the greatest lower bound for aset S ! R if

1. " is a lower bound for S , and

2. if # is a lower bound for S , then " % # .

If a set S has a greatest bound , then we denote it by glb(S).

The greatest lower bound of S is often called the infimum of S ,denoted by inf (S).

Definition: We say that " # R is the greatest lower bound for aset S ! R if

2. if # is a lower bound for S ,

then " % # .

If a set S has a greatest bound

, then we denote it by glb(S).

The greatest lower bound of S is often called the infimum of S ,

denoted by inf (S).

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

Note: glb(S) = 0 # S but lub(S) = 1 &# S .

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

glb(S) = 0 # S but lub(S) = 1 &# S .

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

Note: glb(S) = 0 # S

but lub(S) = 1 &# S .

[0, 1)

Let S = [0, 1).

! lub(S) = 1.

! glb(S) = 0.

Max and Min

Definition:

1. If S contains ! = lub(S), then ! is called the maximum of S(' ! = max(S)).

2. If S contains " = glb(S), then " is called the minimum of S(' " = min(S)).

If S is a finite set with n elements

S = {a1 < a2 < · · · < an},

! an = lub(S) = max(S),

! a1 = glb(S) = min(S).

Max and Min

Definition:

1. If S contains ! = lub(S), then

! is called the maximum of S(' ! = max(S)).

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

1. If S contains ! = lub(S), then ! is called the maximum of S

(' ! = max(S)).

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

2. If S contains " = glb(S),

then " is called the minimum of S(' " = min(S)).

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

2. If S contains " = glb(S), then " is called the minimum of S

(' " = min(S)).

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Max and Min

Definition:

S = {a1 < a2 < · · · < an},

! a1 = glb(S) = min(S).

Empty Set

Problem:

Does every set S that is bounded above have a LUB?

Problem: Is ( bounded?

Claim: $13 is an upper bound for (.Suppose not.

! Then there exists x # ( such that x > $13.

! No such x could exist!!!

! So $13 is an upper bound of (.

Empty Set

Problem: Does every set S that is bounded above have a LUB?

Empty Set

Claim:

$13 is an upper bound for (.Suppose not.

Empty Set

Claim: $13 is an upper bound for (.

Suppose not.

Empty Set

In fact:

! Every real number ! is an upper bound of ( .

! ( is bounded above, but it has no least upper bound.

Empty Set

In fact:

Empty Set

In fact:

Empty Set

We saw that $13 is an upper bound for (.

Claim: 7 is a lower bound for (.

Again, suppose not.

! Then there exists x # ( such that x < 7.

! So 7 is a lower bound of (.! Every real number " is a lower of (.! ( is bounded.

Note: ( is the only subset of R with an upper bound ! and alower bound " with

! < ".

Empty Set

Claim:

7 is a lower bound for (.

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

! So 7 is a lower bound of (.

! Every real number " is a lower of (.! ( is bounded.

! < ".

Empty Set

Again, suppose not.

! So 7 is a lower bound of (.! Every real number " is a lower of (.

! ( is bounded.

! < ".

Empty Set

Again, suppose not.

! < ".

Empty Set

Again, suppose not.

( is the only subset of R with an upper bound ! and alower bound " with

! < ".

Empty Set

Again, suppose not.

! < ".

Axiom: (Least Upper Bound Property or LUBP)

A nonempty subset S ! R that is bounded above always has aleast upper bound.

This Axiom plays a central role in the course!

Important Note

!"lub(S)

If ! = lub(S) and $ > 0 then:

! !$ $ <! =' ! is not an upper bound of S .

! There is an x # S with !$ $ < x " !.

Important Note

!"lub(S)

Note: If ! = lub(S)

and $ > 0 then:

Important Note

!"lub(S)

Note: If ! = lub(S) and $ > 0

Important Note

! "- !#lub(S)

Note: If ! = lub(S) and $ > 0 then:

! !$ $ <!

=' ! is not an upper bound of S .

Important Note

! "- !#lub(S)

Important Note

! "- x S# !$lub(S)

Example

–2 –1 1 2

S = {x # Q | x2 < 2}

' lub(S) =)

2, glb(S) = $)

Example

–2 –1 1 2

S = {x # Q | x2 < 2}

' lub(S) =)

2, glb(S) = $)

Example

!2!2- 0

–2 –1 1 2

S = {x # Q | x2 < 2}

' lub(S) =)

2, glb(S) = $)

Example

!2!2- 0

–2 –1 1 2

S = {x # Q | x2 < 2}' lub(S) =

)2, glb(S) = $

Example

!2!2- 0

–2 –1 1 2

Problem:

What happens to S = {x # Q | x2 < 2} if we only considerrational numbers rather than real numbers?

! S has no lub in Q.

! S has no glb in Q.

Conclusion: Q does not have the Least Upper Bound Property.

Example

!2!2- 0

–2 –1 1 2

Problem: What happens to S = {x # Q | x2 < 2} if we only considerrational numbers rather than real numbers?

Example

!2!2- 0

–2 –1 1 2

Example

!2!2- 0

–2 –1 1 2

Example

!2!2- 0

–2 –1 1 2

Conclusion:

Q does not have the Least Upper Bound Property.

Example

!2!2- 0

–2 –1 1 2

Conclusion: Q does not have the Least Upper Bound Property.Brian Forrest Least Upper Bound Property

Is N bounded?

Problem:

Is N bounded?

Mathematical Induction...October 1, 2010 Brian Forrest Mathematical Induction Principle of...

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